5

Coleges of inierest Igh school senior Include Jway), 8nd Ihat are both expensive and far fron homcuinensye (uton morn Inan 520,O0Q per ycar) bal &7etrom home (...

Question

Coleges of inierest Igh school senior Include Jway), 8nd Ihat are both expensive and far fron homcuinensye (uton morn Inan 520,O0Q per ycar) bal &7etrom home (more Ien 270Kl>Iho studenl decules = seled college Ihat IS nGt oxpensive artd within ?00 mles hoale,how many selectons are possible?If tho studont docidos ahond colleqe thal = not exporsivo and within 200 nule s Irort hoqtle dutinv his first INo yoars collepa,and uut wubuneler enkg tnal @ nocotpcnel butrs Irom home , how many sakch

coleges of inierest Igh school senior Include Jway), 8nd Ihat are both expensive and far fron homc uinensye (uton morn Inan 520,O0Q per ycar) bal &7e trom home (more Ien 270Kl> Iho studenl decules = seled college Ihat IS nGt oxpensive artd within ?00 mles hoale,how many selectons are possible? If tho studont docidos ahond colleqe thal = not exporsivo and within 200 nule s Irort hoqtle dutinv his first INo yoars collepa,and uut wubuneler enkg tnal @ nocotpcnel butrs Irom home , how many sakchons of two collunus ute poss ble?



Answers

Cost of Public Colleges The average cost, in dollars, for tuition and fees for in-state students at four-year public colleges over the period $2000-2014$ can be modeled by the equation$y=4.065 x^{2}+370.1 x+3450$ where $x=0$ comesponds to 2000 $x=1$ corresponds to $2001,$ and so on. Based on this model, for what year after 2000 was the average cost $\$ 8605 ?$ (Source: The College Board, Anmual Survey of Colleges.)

In this video, we're going to be taking a look at using a cumulative frequency graft to estimate percentiles and positions of data values given a percent o. So here I've pre drawn and labelled the graph that we're going to use. And I have expressed the horizontal axis, which is the cost in thousands of dollars just for ease of writing the numbers and are vertical accesses are cumulative frequency. So in the table were given the individual frequencies of the General and we're gonna accumulate them or add them up as we go. So in the first interval between 3000 and 4000 um, we have five. That means we're gonna be starting out our graph at zero, because before 3000 we don't have anything lower than 3000 and then between 3000 and 4000 we have five, so that accumulates air adds up 25 In the next interval, we have an additional six. So between 4000 and 5000 we have an additional six data values That takes us up to 11 between 5000 and $6000. We have an additional 18 which takes us up 2 29 these air estimations because on a graph, when we're dealing with numbers of these scale, the best we can do is make an estimate between 6000 and 7000. We have an additional 24 data values. So that brings our grand total are accumulated total up to 53. So about right there and then between seven and 8000 we have another 19 which takes our cumulative total to 72. So right about there between 8000 and 9000 we have an additional eight, which takes our cumulative total up to 80 and between 9000 and 10,000. We have an additional five, which takes our grand total up to 85. So we know that our grand total that told remember pieces of data that we're talking about here is 85. So now we want to try a Z best we can to connect these dots. What we're assuming is that the data is evenly distributed. Those five data values are evenly distributed between 3000 and 4000. Same thing for each of these other individuals. So we connect with a straight line to show that we are assuming on approximately even distribution. Um, given this format and, um, the whiteboard. I'm doing the best that I can to try to make each other's line segments as straight as possible. All right, so now what we're gonna be doing is using this graph toe estimate some percentiles. So for question A, we are asked to find the 30th percentile. So we needed a little a review of a formula. So we have this formula that says that C is equal to end times p over 100 where n is the number of data values that were dealing with so 85 times the percentile 30 over 100. And this 30 over 100 really just serves to convert Teoh per percentage for us. So we've got 85 and I want 30% of that. So 30 over 100. Mathematically, this is the same as multiplying 0.3 times 85 which for me is a little bit easier to wrap my head around If I can come forth 30th percentile to 30% and multiply it by 85 eso I'm looking for the data value that represents the lowest 30% of the data. This calculation comes out to be 25 point five, which, if I round up, then if there's 25 a half numbers less than that, I'm looking for the 26 data value. Okay, so I'm looking for where my My total frequency is 26. So if I come over here on my cumulative frequency chart toe about where 26 is and that's a ride about here, right, that's about 26 just above that line, I'm gonna estimate by coming over here trying to draw a straight of a line as I can. I'm using kind of a dash line and then follow that line down to my data value. This is my estimate. So it looks like it's pretty close to 6000. So my estimate here will be between 11,000 pretty close to 6000. So I might say something like 5875. All right, so for question be it wants to know the 50th percentile following a similar procedure there. I know 50th percentile. 50% means I'm looking for 50% of 85. So this 0.5 takes care of the dividing by 150% of 85 is 42.5. So that means I'm looking for the 43rd number, the 43rd date of value. So I'm gonna come over here again. Here's 40. Here's 50. Right in the middle is 45. So 43 is going to be in about the middle of that range. So if I come over to my Graf again, um, trying to stay in the middle to until I hit the graph and then follow that line down. So that looks to be pretty close to about that third line over maybe a little bit less than that. So I'm gonna estimate that that's going to be about 6670. So your answers might be just a little bit off, depending on how you draw on your graph. But these are all pretty close. These are pretty good estimates, Part C. We want to know the 75th percentile, so I'm gonna multiply 0.75 times 85 which gives me 63.75. In other words, I'm looking for my 64th data value. So 64th day, the value is gonna be just below this line right here. So if I come over here staying kind of just below that line, and then I'm gonna follow this till I hit the graph and then wanna follow that down to the data. So that looks to be about 7600. I'm using my approximately equal to because thes are estimations were making finally, for finding the percentiles. I'm looking for the 90th percentile. So 0.9 times 85 gives me 76.5, which then means I'm looking for the value associated with the number in the 77th position. So 70 75 77 is gonna be about the middle here. We're gonna follow this over to the graph. I got my middle, got a little bit off key that off track about here, and then I'm gonna follow that, Dale. So that looks to be a little bit over 8600. So well, approximate. That is like, I don't know. Let's say 86 50. Okay, so those are all estimations where you start by finding the value the what? Which data? Which position you're looking for and then following that over to the graph and then down to the data value. Okay, the next part of the question we're looking at is going to be working in the reverse. If we're given a data value, we're going to try to identify what percentile that it's. And I'm gonna make this graph here a little bit larger. So I took a similar, um, cumulative frequency graph, Same same shape. You can see, uh, that I drew earlier, and we're gonna use this graph to work in the other direction. So again, we're going to need to have a formula that we're working with and from our textbook. We have this information that tells us that the percentile which I'll call P, is the number of values lower than our data value. Um, times the number values, lower time, Earth. Sorry, not times, but plus 0.5. And then we divide by the total number of numbers that we have, which in this case is 85 and then we multiplied by 100. So this turns it into ah, decimal value. And then multiplying by 100 turns out decimal value into a percent. So if I confined which position my data value is in, then I can figure out how many or lower than that. That's what this number represents. At 1000.5, divide by 85 then turn that back into a percent. So for Part E, we are given the data value of 5500 we want to know what percentile that represents. So these are costing thousands. So 5500 is going to be about halfway between 5000 and 6000. So we're gonna work in reverse of what we did before 5500 approximately here. So now we're gonna move up to the graph from there, and it looks like it's right about here, right? At about 20 if I come over and look here. So that means if I'm at the 20th data value, there are 19 data values that are lower than that. So this is our number sign. This is 19 values lower. We're gonna add 190.5 to it, divide by 85 and multiply by 100. So 19.5, divided by 85 comes out to be 22.9. So that means that we are approximately looking at the sorry. A little ahead of myself there. 23rd percent O Okay, So question ask says, Okay, what if our data value is 7200? Um, what if our cost of $7200 so again, we're going to start at 7200? So that's gonna be about here, right? Somewhere between 7000 and 7300. And I'm just gonna try Sava and see how that works. Follow that up and then follow that across. So 7200 looks like it's about at 55 um, 56 somewhere in there. So let's say that this is the 56 value, which means that there are 55 data values lower. We're gonna add 0.5, divide by 85 multiply by 100. So 55.5, divided by 85 and multiplied by 100 gives us about 65.29 So this isn't about the 65th just using normal rounding about the 56% T per. Oops, I spelt that entirely wrong. Let's try that again. Percentile. Okay. Um so to to go part G, we're looking for the percentile associated with the cost of $6500. So $6500 is going to be like right smack dab in the middle between 6000 and 5000. So if we come up here so it looks like right about in here, um, and we move over, looks to me like it's it about the 43rd position or so. So that means we have 42 numbers less than that, 42.5 out of 85 times 100 then that becomes the 50th percentile. And the last one h we want the percentile for the data value of 8300. So 8003 hundreds about here, once chemical upto our graph. And we'll follow that over trying to make my lines of straight as I can. Um, and that looks to be like about the 75th value, which means there are 74 lower than that. So 74 plus 740.5 that's 74.5 divided by 85 and multiply by 100. Which means then that this is at about the 89th percentile

So in this question, we're gonna be taking a look at some costs and trying to estimate percentiles and estimate values based on percentiles using a cumulative frequency or a percentile graph. So I have the data here, the frequency table, and we're going to find the humility frequency, the and the relative frequency, which is just a fancy way of saying percent and then make a percent autograph to answer these questions. So this is just fancy way of saying percentages and I'm gonna write These is decimals because I've expressed my relative frequency has here is decimals where 0.1 is 10% 20% 30% all the way up toe one, which represents 100%. And then my horizontal axis. The cost is three, which represents 3000.5. Um, I didn't wanna have to write all those really long numbers out, so I just abbreviated and expressed the cost in thousands of dollars. All right, so let's first start by completing our cumulative frequency table. So the data that's in this range there's five data values between those two numbers, So the total there is five, and then a cumulative frequency graft means you're just going to keep adding them up. So they're six more data values that are between the end of this interval on the end of this interval. So from 4000.5 to 5000.5 for a grand total of 11 keep adding get 29 29 plus 24 is 53 53 plus 19 72 72 plus eight is 80. An 80 plus five is 85. So we get our grand total there of 85 then to find a relative frequency, we're gonna take each of the frequencies and divide by 85. So what part of the overall distribution is five, So five out of 85 is 850.60 point 068 or 6.8%. So I'm gonna write it as a decimal. So 0.68 and then 11 out of 85 is 850.13 13%. 29 out of 85 is 34 point month, 34.1% so 0.341 53 out of 85 is 62.3% or points 6 to 3, 72 out of 85 is 84.7% or 0.847 80 out of 85 94.1% some 0.9 for one and then 85 out of 85 100% or just one. So now we're gonna use this column, this relative frequency comb to make our relative frequency graph. And so the were using the end of the interval to mark where we've accumulated too. So when we start to make this graph, we don't have anything when we start off. So at three, we have zero. There is nothing. We haven't accumulated anything. But by the time we get to three, when we get to 4000 we've accumulated 6.8% So each line is representing 5%. So I'm a little bit above that line by the time I get to 5000 and at 13% by the time I get to 6000 I'm at 34.1%. So a little bit below 35% by the time I get to 7000. I'm at 62.3% by the time we get to 8000 or 84.7% by the time we get to 9000. Were at 94.1%. And by the time we get to 10,000 more at 100%. So this is our cumulative frequency graph. I'm gonna attempt to draw a straight of a long as I can between each of these thoughts to suggest that the data is evenly distributed between those intervals. So a little bit hard to dio using that, um, this mechanism, this white board, but did the best we can. Okay, so this is our cumulative frequency curve. This is a little bit bad right here. Straighten it out a little bit. Okay? And now we can use this to make some estimations about what the percentiles are for Any of the costs that are between 3000 and 10 thousands. So our first question part A says, Okay, we have the 30th percentile. What cost is associated with the 30th percentile? So we start with our given information. We know 30th percentile means 0.3. So we're gonna use this graph here. Ever try to go a straight as we can over to the graph following it in this direction. And then once we get here, we're going to estimate what that cost is by following the line down. So this is a pretty big gap, but we can estimate the best of our ability so that we know this is 5000. This is halfway. This is pretty close to 6000. So something approximately a cost of approximately Ah, 5800 is a reasonable estimate. Okay, let's look at the next percent up. 50th percentile. So again, we're gonna come over here to the 50th percentile, and we're gonna follow from that percent out over two or graph in this direction. And then we're gonna follow that one down vertically to our data value. And this looks to be, by this estimate, pretty close to about 6500. That seems like a good estimate. Okay, Part C, we're looking at the 75th percentile, so the 75th percentile is gonna be halfway between 70 and 80%. And we're going to follow the graph over again, going from that percent out to the graph, and then we're gonna follow that down and a little bit o a little bit. Well, we let me try that again. Looks like we're a little bit above this line. So about here. So a little over 75 100. So let's say on a 7600 seems reasonable. And then our last percentile we're going to estimate is the 90th percent o so following over again from the 90th percentile over to the graph and then following that down, making the lines a straight as I can. So a little over 8500 about halfway. So let's say 8750 as an estimate. Okay, so now we want to go the other direction. So we're going to be given the estimate of the costs, and we want to find what percentile is associated with that. So for Part E, we're looking at a cost of 5500. So what percent? I would be associated with that cost. So we're gonna work in reverse. We're gonna go over to that cost of 5500. Instead of going from the frequency to the graph, we're going to go the other direction, going to go from the data to the graph, and then we're gonna pull that over to the frequency and estimate, and this looks to be at about the 25th percentile f We're looking for the cost of 7200. What percentile is associated with that? So 7200 is gonna be right about here and again. We're gonna follow that up to the graph moving up in a vertical direction from our horizontal access to our relative frequency graph, and then we're gonna pull that over, and that looks to be about the 65th percentile G. We're looking at a cost of 6500. So 6500 looks like we kind of estimated that. So we can kind of work this one backwards. Right? So this direction here, um, that is associated with about the 50th percentile and then h is a cost of 8300. And so 8300 would be a little bit past the middle here right around there. And we'll follow that as best we can to the graph and then over. And that looks to be approximately the 85th percentile. So these are estimates. When we have this graph, we have an interval. We're making an estimate. We're making a lot assumptions about drawing straight lines, but to the best of my drawing ability here, these are reasonable estimates.

Now we're going to talk about the increase of college expenses versus inflation. So we're going to take the College of Harvard as an example and we're just going to look at their Uh tuition prices in 2000 2010 and 2020. So first I want to calculate the percentage increase. So let's first calculate 2000 to 2010. So that's going to be, So this is going to be a roughly 52 8% increase. So that's for 2000 To 2010. Now it's calculate 2010- 2020. She did me put that further away And this one is to be roughly around 41.6%. So between 2000 2010 we had a 52 increase in price or 52.8 increase in price. And the following decade after that, the increase was 41.6. And so these numbers are extremely large compared to the inflation rate over the course of these periods in the United States. So this means that inflation is not the reason why college expenses has been increasing. It is due to other reasons.

All right. I'm re recording this video because I'm not sure it recorded correctly the first time. Um So all the information is already here, uh to answer part A um What I did is I looked at the graph and identified where the slope was the greatest and that slope is the greatest around uh 1985 to 1986. And that is for the private colleges. That's the blue line in part B. I looked at the red line and look for the greatest slope. I found the greatest slope to be in the 2003 to 2000 and four school year. Um Now, for the last part, yeah, we're taking inflation back out, which is why I'm dividing by 1.3 So I'm dividing by 100 and 3%. Um Taking inflation back out of the numbers for the 1975 school year. So reading off the graph, I got $13,500 for private colleges and $6000 for public colleges in those years, just divide both of those by 1.3 to the 33rd power because that's 33 years from 1975 to 2000 and eight is 33 years. So if I do that, I get 5000 $100 for the private college and $2300 for the public college.


Similar Solved Questions

5 answers
QuestionOjines#tch Pulun humars Delorz?AfIuArhicood; UdllusachordaleQueticn"5puin3NEch DNnCeDuvtamDSot'D *Opatt0 u47iType hereucLLESC
Question Ojines #tch Pulun humars Delorz? AfIu Arhicood; Udllusa chordale Queticn "5puin3 NEch DNnCe Duvtam DSot 'D * Opatt0 u47i Type here uc LL ESC...
5 answers
Determine the limit (answer as appropriate, with number; €_ ~or does not erist).1-29 lim 4 1 - 6r* + 9
Determine the limit (answer as appropriate, with number; €_ ~or does not erist). 1-29 lim 4 1 - 6r* + 9...
5 answers
Jnswict Uti Inten hotation | DartIc speedina 05? (Euter YcuiWnen< [ < %,siowIno down? (Ertu Your onswver Using Interval notation )Wmen{or
Jnswict Uti Inten hotation | DartIc speedina 05? (Euter Ycui Wnen < [ < %, siowIno down? (Ertu Your onswver Using Interval notation ) Wmen{or...
5 answers
Fropiem(ZU points) (Similar t0 problem 9.8 For each of the two functions below , dulermne the system gradient 3Y SUCL Hamiltonian system find the COrTespondling gradient or Hamil- toniau function. sketch the phase portrait of the ~at1 =I* 2rv, V =y ZIy(b) r =r Zcy.y
Fropiem (ZU points) (Similar t0 problem 9.8 For each of the two functions below , dulermne the system gradient 3Y SUCL Hamiltonian system find the COrTespondling gradient or Hamil- toniau function. sketch the phase portrait of the ~at 1 =I* 2rv, V =y ZIy (b) r =r Zcy.y...
5 answers
22.13 he graph of function f is given. Use the graph to find each ol the following; The numbers, If any, at which f has relative maximum What are thesu Ialive maxima? The numbers. If any: at which has relative minimum. What are these relalive minima?TutoringQuestonmplala tour
22.13 he graph of function f is given. Use the graph to find each ol the following; The numbers, If any, at which f has relative maximum What are thesu Ialive maxima? The numbers. If any: at which has relative minimum. What are these relalive minima? Tutoring Queston mplala tour...
1 answers
Solve each linear programming problem by the simplex method. $$ \begin{array}{cc} \text { Maximize } & P=5 x+3 y \\ \text { subject to } & x+y \leq 80 \\ 3 x & \leq 90 \\ x & \geq 0, y \geq 0 \end{array} $$
Solve each linear programming problem by the simplex method. $$ \begin{array}{cc} \text { Maximize } & P=5 x+3 y \\ \text { subject to } & x+y \leq 80 \\ 3 x & \leq 90 \\ x & \geq 0, y \geq 0 \end{array} $$...
5 answers
9 frictiohes 8 Acrale ol mass What Is the specd force of 10 N along the incline. at the bottom of the 1 U at the top of the wilh incline? 1 Initial vclocily hthen horizontshed ms [s H 10rn hnetice2.3 nstRmis
9 frictiohes 8 Acrale ol mass What Is the specd force of 10 N along the incline. at the bottom of the 1 U at the top of the wilh incline? 1 Initial vclocily hthen horizontshed ms [s H 10rn hnetice 2.3 nst Rmis...
1 answers
Express each temperature in degrees Fahrenheit. $$75^{\circ} \mathrm{C}$$
Express each temperature in degrees Fahrenheit. $$75^{\circ} \mathrm{C}$$...
5 answers
What is the pOH of a solution when [OH] = 1.70x 10-5M?0 10.200170D0.225.204.77
What is the pOH of a solution when [OH] = 1.70x 10-5M? 0 10.20 0170 D0.22 5.20 4.77...
5 answers
Determire if tle zequenc2 converges of diverge If the gequenc2 corverges , fnd it3 limit Tbe wczk akculd be done algebraically Xcu can Yjuf calcuator zd yoL zhould, to check tlat youz angwez correct, but tke work zhculd be algebzaic. 4n3 3n' + 5V2n'(L)Z(C.8)n Sn =1+
Determire if tle zequenc2 converges of diverge If the gequenc2 corverges , fnd it3 limit Tbe wczk akculd be done algebraically Xcu can Yjuf calcuator zd yoL zhould, to check tlat youz angwez correct, but tke work zhculd be algebzaic. 4n3 3n' + 5 V2n' (L)Z (C.8)n Sn =1+...
5 answers
Question 34ptsWhich of the following ions/atoms would have the smallest radius?
Question 34 pts Which of the following ions/atoms would have the smallest radius?...
5 answers
Note: is selecled Qucstion L What What is A production process Use the binomial probability 241 the standard Ji produces deviation that that 2%/ Tuncnuu the sample 1 cuntams sxacdlyr Io sample of $ parts less of the sampld 1| 1
Note: is selecled Qucstion L What What is A production process Use the binomial probability 241 the standard Ji produces deviation that that 2%/ Tuncnuu the sample 1 cuntams sxacdlyr Io sample of $ parts less of the sampld 1| 1...
5 answers
1/2 POINTSPREVIOUS ANSWERSSERCP11 10.4.OP.033.MY NOTESASK YOUR TEACHERHow many molecules are present In sample of an ideal ga5 that occupies volume of 2.60 cm? , Is at temperature of 2091 and at atmospheric pressure? 6.5E19 moleculesHow many molecules of the gas are present the volume and temperature are the same as good vacuum)?part (a), but the pressure now .50 * 10-11 Pa (an extremelymolecules
1/2 POINTS PREVIOUS ANSWERS SERCP11 10.4.OP.033. MY NOTES ASK YOUR TEACHER How many molecules are present In sample of an ideal ga5 that occupies volume of 2.60 cm? , Is at temperature of 2091 and at atmospheric pressure? 6.5E19 molecules How many molecules of the gas are present the volume and temp...
5 answers
Total25212. The following data were obtained from an inde- pendent-measures research study comparing three treatment conditions. Use an ANOVA with & 05 to determine whether there are any significant mean differences among the treatments:Treatment52 6 2 3 5 3 03 2 4 ; 52 3 0 2
Total 252 12. The following data were obtained from an inde- pendent-measures research study comparing three treatment conditions. Use an ANOVA with & 05 to determine whether there are any significant mean differences among the treatments: Treatment 5 2 6 2 3 5 3 0 3 2 4 ; 5 2 3 0 2...
5 answers
(Vandermonde Matrices) Symbolically define the variables through 1 and then find the determinant of the following matrices a2 a3 62 63 b c2 c3 d2 Hint: Use simplify. * Use part(a) to compute the determinant ofF =-1 Remark: Here, i is the complex number such that i2 =_
(Vandermonde Matrices) Symbolically define the variables through 1 and then find the determinant of the following matrices a2 a3 62 63 b c2 c3 d2 Hint: Use simplify. * Use part(a) to compute the determinant of F = -1 Remark: Here, i is the complex number such that i2 =_...

-- 0.059239--